Console.log() is passed a reference to the object, so the value in the Console changes as the object changes. To avoid that you can: console.log(JSON.parse(JSON.stringify(c))) MDN warns: Please be w... Read More
Yes, you can. $a = $b = $c = $d = array();... Read More
How they are stored is an implementation detail (depends on the compiler). For example, in the GCC compiler, on most machines, read-only variables, constants, and jump tables are placed in the text s... Read More
Try something like this: SET MY_PATH=C:\Folder with a space "%MY_PATH%\MyProgram.exe" /switch1 /switch2... Read More
When you use a command substitution (ie the $(...) construct), you are creating a subshell. Subshells inherit variables from their parent shells, but this only works one way - a subshell cannot modif... Read More
Surround that select with parentheses. SET @v1 := (SELECT COUNT(*) FROM user_rating); SELECT @v1;... Read More
Copied directly from http://www.lighthouse3d.com/tutorials/glsl-tutorial/data-types-and-variables/. The actual site has much more detailed information and would be worthwhile to check out. Variable... Read More
I've been looking for better way of doing it recently. Associative array sounded like overkill for me. Look what I found: suffix=bzz declare prefix_$suffix=mystr ...and then... varname=prefix_$suffi... Read More
Between those two syntaxes, you should really choose the one you prefer :-) Personally, I would go with your second solution in such a case (Variable interpolation) , which I find easier to both writ... Read More
You can use the typeid operator: #include <typeinfo> ... cout << typeid(variable).name() << endl;... Read More